Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Section 4.2 - The Natural Exponential Function - 4.2 Exercises - Page 341: 17

Answer

See the image below.

Work Step by Step

(a) To make our work easier we can first graph $y=e^x$ and then apply transformations to get the graph we are asked for. (You can see this graph, a parent graph, as light grey graph in the image). Next we can get the graph of $y_1=\frac{e^x}{2}$ by shrinking the graph of the parent function vertically by the factor of $2$. (Shown as the blue graph in the image) Now we can simply get $y_2=\frac{e^{-x}}{2}$ by reflecting $y_2$ about $y$-axis. (Red graph in the image) Using the graphical addition (Explained in the section 2.7) we can sketch the graph of $cosh(x)=\frac{e^x+e^{-x}}{2}$ using the graphs of $y_1$ and $y_2$. (b) To prove that $cosh(x)=cosh(-x)$ we can input both of these values and then compare them. (For a better understanding we can follow the same way we did in (a)). $cosh(x)=\frac{e^x+e^{-x}}{2}$ $cosh(-x)=\frac{e^{-x}+e^{-(-x)}}{2}=\frac{e^{-x}+e^{x}}{2}$ It's clear that both of them have the same expression which means the same graph and so they are equal to each other.
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